### POLONICI MATHEMATICI LXIX.1 (1998)

**Hammerstein equations**

**with an integral over a noncompact domain**

## by Robert Sta´ nczy ( L´od´z)

**Abstract.** The existence of solutions of Hammerstein equations in the space of bounded and continuous functions is proved. It is obtained by the Schauder ﬁxed point theorem using a compactness theorem. The result is applied to Wiener–Hopf equations and to ODE’s.

## 1. Introduction. In 1929 Hammerstein published paper [4] concerning nonlinear integral equations with integral operators of the form

## (1) Sx(t) =

\

T

## G(t, s)r(x(s), s) ds.

## If the domain of integration is compact and the functions G and r are con- tinuous, then equation (1) can be considered in the space C(T ) of continuous functions. It can also be considered in the space L

^{p}

## (T ) of integrable func- tions (where T has a ﬁnite or an inﬁnite measure).

## In this paper, a noncompact domain of integration is considered. We work in the space of bounded and continuous functions. Therefore, to apply the Schauder theorem, we need suﬃcient conditions for compactness in this space. Some of them are presented in [1]. In our paper, however, necessary and suﬃcient conditions are stated. The main theorem gives conditions for the complete continuity of the Hammerstein operator. It is the extension of the results of [5] where T = (−∞, ∞). Some applications to the Wiener–

## Hopf equation on a half-line or on a half-space are given. The theorem can also be applied to ODE’s of second order with boundedness as a boundary condition since the Green function is continuous.

## 2. Preliminaries and auxiliary lemmas. Let X be a Banach space with norm k · k and let T be a locally compact metric space countable at

*1991 Mathematics Subject Classification: 45G10, 47H30.*

*Key words and phrases* : Hammerstein operator, Wiener–Hopf equation.

[49]

## inﬁnity, i.e. T = S

∞i=1

## K

i## where (K

i## )

i∈N## are compact and satisfy K

i## ⊂ int(K

i+1## ) for any i ∈ N.

## For V ⊂ X let conv V denote the smallest convex, closed set of X con- taining V . We write K ⋐ T if K ⊂ T is compact.

## We shall also use the following function spaces:

## • L(X) — the space of bounded linear operators from X to X,

## • BC(T, X) — the space of bounded continuous functions from T to X with sup norm k · k

_{∞}

## ,

## • C(K, X) (where K ⋐ T ) — the space of continuous functions with sup norm denoted as above,

## • CC(X, X)−the space of completely continuous functions from X to X, i.e. continuous and compact (mapping bounded subsets of X into relatively compact ones), with metric

## d(f, g) = X

∞ n=1## 2

^{−n}

## sup

_{kxk≤n}

## kf (x) − g(x)k 1 + sup

_{kxk≤n}

## kf (x) − g(x)k .

## The space CC(X, X) thus deﬁned is a Fr´echet space, i.e. locally convex and completely metrizable. The convergence in this space is the uniform convergence on bounded subsets of X.

## For y ∈ X and x ∈ BC(T, X), we write lim

t→∞## x(t) = y if

## ∀

ε>0## ∃

K⋐T## ∀

t∈T \K## kx(t) − yk < ε

## or one can understand it in terms of the Alexander compactiﬁcation of the space T with the point ∞.

## We shall also use the Bochner integral over the domain T . Let µ denote a measure on B(T ) (the σ-algebra of Borel subsets of T ), ﬁnite on compact subsets of T . For the Bochner integral of vector-valued functions see [3, pp.

## 44–52] or [6, pp. 132–136].

## Let us recall a property of the Bochner integral to be used in the sequel.

## Theorem 1. Let x : T → X be a Bochner integrable function. Then, for each set K ⊂ T of finite measure, we have

\

K

## x(t) dt ∈ µ(K) conv(x(K)).

## P r o o f. See [3, Corollary 8, p. 48].

## Now, we prove some auxiliary lemmas concerning the Nemytski˘ı opera- tor. Let r : X × T → X.

## Lemma 1. Suppose that the mapping T ∋ t 7→ r(·, t) ∈ CC(X, X) is well defined and continuous. Then r : X × T → X is continuous.

## P r o o f. Fix t

1## ∈ T , x

1## ∈ X and ε > 0. There exist δ, γ > 0 such that

## d(t, t

1## ) < δ implies d(r(·, t), r(·, t

1## )) < ε/2, whence kr(x, t

1## ) − r(x

1## , t

1## )k

## < ε/2 for any kx − x

1## k < γ. Then, for kx − x

1## k < γ and d(t, t

1## ) < δ, we get kr(x, t) − r(x

1## , t

1## )k ≤ kr(x, t) − r(x, t

1## )k + kr(x, t

1## ) − r(x

1## , t

1## )k

## ≤ ε/2 + ε/2 = ε.

## Lemma 2. Suppose r : X × T → X is continuous. Let K ⋐ T , x ∈ C(K, X) and ε > 0. Then there exists δ > 0 such that, for any t ∈ Kand y ∈ C(K, E) satisfying ky(t)−x(t)k < δ, we have kr(y(t), t)−r(x(t), t)k < ε.

## P r o o f. Suppose that, on the contrary, there exist x

0## ∈ C(K, X), ε

0## > 0, t

n## ∈ K and y

n## ∈ C(K, X) such that ky

n## (t

n## ) − x

0## (t

n## )k < 1/n and

## kr(y

n## (t

n## ), t

n## ) − r(x(t

n## ), t

n## )k ≥ ε

0## .

## From the sequence t

n## ∈ K we can extract a convergent subsequence t

_{n}

_{k}

## → t

0## , whence x

_{0}

## (t

_{n}

_{k}

## ) → x

_{0}

## (t

_{0}

## ) and ky

_{n}

_{k}

## (t

_{n}

_{k}

## ) − x(t

_{n}

_{k}

## )k < 1/n

_{k}

## . Hence y

nk## (t

nk## ) → x

0## (t

0## ). Thus r(y

nk## (t

nk## ), t

nk## ) − r(x

0## (t

nk## ), t

nk## ) → 0, which con- tradicts our assumption.

## Lemma 3. Let the mapping T ∋ t 7→ r(·, t) ∈ CC(X, X) be continuous.

## Then, for any M > 0 and any K ⋐ T , the set {r(x, t) : kxk ≤ M, t ∈ K}

## is relatively compact.

## P r o o f. Fix M > 0. By the complete continuity of r(·, t) for any t ∈ K, we can choose an ε/2-net (r(x

^{t}

_{i}

## , t))

^{k}

_{i=1}

^{t}

## of the set {r(x, t) : kxk ≤ M } . Deﬁne e r : T → CC(X, X) by e r(t) = r(·, t). Since e r|K is uniformly continuous, we may choose δ > 0 such that d(s, s

^{′}

## ) < δ implies d(e r(s), e r(s

^{′}

## )) < ε/2.

## Let s

1## , . . . , s

p## be a δ-net of K. Then (r(x

^{s}

_{i}

^{j}

## , s

j## ))

^{k}

_{i=1}

^{sj}

^{p}

_{j=1}

## is an ε-net of the set {r(x, t) : kxk ≤ M, t ∈ K}. Indeed, for any s ∈ K and kxk ≤ M , we may choose s

_{j}

## such that d(s, s

_{j}

## ) < δ and x

^{s}

_{i}

^{j}

## such that

## kr(x

^{s}

_{i}

^{j}

## , s

_{j}

## ) − r(x, s

_{j}

## )k ≤ ε/2.

## Then

## kr(x

^{s}

_{i}

^{j}

## , s

_{j}

## ) − r(x, s)k ≤ kr(x

^{s}

_{i}

^{j}

## , s

_{j}

## ) − r(x, s

_{j}

## )k + kr(x, s

_{j}

## ) − r(x, s)k

## ≤ ε/2 + ε/2 = ε.

## Lemma 4. Let M > 0. Suppose that the mapping T ∋ t 7→ r(·, t) ∈ CC(X, X) is continuous, and that there exists b ∈ BC(T, X) such that

t→∞

## lim sup

kxk≤M

## kr(x, t) − b(t)k = 0.

## Then r(B(0, M )× T ) is bounded in X (B(0, M ) is the closed ball with centre 0 and radius M > 0).

## P r o o f. Fix ε > 0 and choose K ⋐ T such that kxk ≤ M , t 6∈ K imply

## kr(x, t) − b(t)k ≤ εkr(x, t)k ≤ ε + kb(t)k.

## By Lemma 3, T

K,M## = sup

_{kxk≤M,t∈K}

## kr(x, t)k < ∞, whence kr(x, t)k ≤ max{ε + kbk

∞## , T

K,M## }.

## 3. Main results

## Theorem 2. Let T be a metric, locally compact space countable at ∞ and let X be a Banach space. Then the relative compactness of the set F ⊂ BC(T, X) is equivalent to the conjunction of three conditions:

## 1

^{o}

## The set {x(t) : x ∈ F } is relatively compact in X for each t ∈ T . 2

^{o}

## For each K ⋐ T the functions in F

K## = {x|K : x ∈ F } are equicon- tinuous.

## 3

^{o}

## For each ε > 0, there exist δ > 0 and K ⋐ T such that, for any x, y ∈ F , if kx|K − y|Kk

∞## ≤ δ, then kx − yk

∞## ≤ ε.

## P r o o f. Necessity. If F ⊂ BC(T, X) is relatively compact then the Ascoli–Arzel`a theorem implies 1

^{◦}

## and 2

^{◦}

## . If 3

^{◦}

## were not satisﬁed, there would be ε

0## > 0 and sequences (x

n## ), (y

n## ), (K

n## ) such that

## kx

n## |K

n## − y

n## |K

n## k

∞## ≤ 1

## n and kx

_{n}

## − y

n## k

∞## > ε

_{0}

## .

## We may extract a convergent subsequence x

_{n}

_{k}

## − y

nk## → x − y. But then x|K

n## = y|K

n## for each n ∈ N, which contradicts kx − yk

∞## ≥ ε

0## .

## Sufficiency . Take ε > 0 and choose K ⋐ T from condition 3

^{◦}

## . By the Ascoli–Arzel`a theorem the set {x|K} has a ﬁnite δ-net: {x

1## |K, . . . , x

l## |K}.

## Then, from 3

^{◦}

## we see that {x

_{1}

## , . . . , x

_{l}

## } is an ε-net for F .

## Remark 1. The above theorem, with a similar proof, can be extended to the case of X being a complete, metric space.

## Remark 2. For T = [0, ∞), some suﬃcient conditions for compactness in BC([0, ∞) are stated in [1] with the use of measures of noncompactness.

## Remark 3. Condition 3

^{o}

## is satisﬁed if, for any x, y ∈ F ,

t→∞

## lim kx(t) − y(t)k = 0.

## Example. Let T = X = R, F = {sin(x), cos(x)}. Then F is compact in BC(T, X) though the condition of Remark 3 is not satisﬁed.

## Theorem 3. Let T be a metric, locally compact space countable at ∞ and let X be a Banach space. Suppose T is equipped with a measure on the σ-algebra of Borel subsets of T , finite on compact sets. Define an integral operator S on BC(T, X) by

## Sx(t) =

\

T

## G(t, s)r(x(s), s) ds.

## Assume that

## (i) G : T × T → L(X) is continuous, (ii) sup

_{t∈T}

T

T

## kG(t, s)k ds < ∞, (iii) ∀

_{K⋐T}

## lim

_{t→∞}

T

K

## kG(t, s)k ds = 0, (iv) ∀

_{ε>0}

## ∀ b

^{K⋐T}

## ∃

K⋐T## ∀

_{t∈}

^{K}

## b

T

T \K

## kG(t, s)k ds < ε, (v) the map t 7→ r(·, t) ∈ CC(X, X) is continuous,

## (vi) there exists a function b ∈ BC(T, X) such that, for each M > 0,

t→∞

## lim sup

kxk≤M

## kr(x, t) − b(t)k = 0.

## Then S maps BC(T, X) into BC(T, X) and is completely continuous.

## P r o o f. Deﬁne L = sup

_{t∈T}

T

T

## kG(t, s)k ds. By (ii), L < ∞. The as- sumptions on r imply its boundedness. Hence we get the existence of the integral

T

T

## G(t, s)r(x(s), s) ds and the boundedness of the operator S on any bounded set.

## To get the continuity of Sx, we divide T into a compact set K and apply the continuity of the function G there, and the noncompact set T \ K where we use (iv). Similarly, using (vi) on the noncompact set and (v) on the compact one, we estimate the norm kSx(t) − Sy(t)k. Taking into account (ii) we get the continuity of the operator S.

## Now, we show that S is compact, i.e. for each M > 0, the set S(B(0, M )) is relatively compact. In order to use the compactness criterion in BC(T, X), we have to show that conditions 1

^{o}

## –3

^{o}

## are satisﬁed.

## 1

^{o}

## Fix t

0## ∈ T and ε > 0. We shall ﬁnd a ﬁnite ε-net for A := {Sx(t

0## ) : kxk < M }. Choose K ⋐ T such that kr(x, t) − b(t)k ≤ ε/(2L) for kxk < M and t ∈ T \ K. Set x

0## =

T

T \K

## G(t

0## , s)b(s) ds. Consider B := n

^{\}

K

## G(t

0## , s)r(x(s), s) ds : kxk

∞## ≤ M o .

## From Lemma 3 we see that the set Z

t0## := {G(t

0## , s)r(x, s) : s ∈ K, kxk ≤ M } is relatively compact in X. But the integrals in B belong to the convex hull µ(K) conv Z

t0## , so by the Mazur theorem, there exists a ﬁnite ε/2-net x

1## , . . . , x

p## of B. We have thus obtained an ε-net x

0## + x

1## , . . . , x

0## + x

p## of A.

## Indeed,

## kSx(t

0## ) − (x

_{0}

## + x

_{j}

## )k ≤

\

K

## G(t

_{0}

## , s)r(x(s), s) ds − x

_{j}

## +

\

T \K

## kG(t

0## , s)k · kr(x(s), s) − b(s)k ds

## ≤ ε 2 + L ε

## 2L = ε.

## 2

^{◦}

## Let ε > 0 and b K ⋐ T . Choose, by (iv) and (vi), K ⋐ T such that kr(x, s) − b(s)k ≤ kbk

∞## ,

## whence

## kr(x, s)k ≤ 2kbk

∞## and

^{\}

T \K

## kG(t, s)k ds ≤ ε/(8kbk

∞## ) for any t ∈ b K.

## Moreover, using the uniform continuity of G on b K × K, we can choose some δ > 0 such that, for any t, t

_{1}

## ∈ b K with d(t, t

_{1}

## ) < δ we have kG(t, s) − G(t

1## , s)k ≤ ε/(2µ(K)T ). Then

## kSx(t) − Sx(t

1## )k

## ≤

\

K

## kG(t, s) − G(t

1## , s)k · kr(x(s), s)k ds

## +

\

T \K

## kG(t, s) − G(t

1## , s)k · kr(x(s), s)k ds

## ≤ ((µ(K) · ε)/(2µ(K)T )) · T + (2ε/(8kbk

∞## )) · 2kbk

∞## = ε.

## 3

^{◦}

## Let ε > 0. From (vi), we can choose K ⋐ T such that, for any s ∈ T \K and kxk ≤ M we have kr(x, s) − b(s)k ≤ ε/(4L). Moreover, from (iii) we can choose b K ⋐ T such that

T

K

## kG(t, s)k ds ≤ ε/(4T ) for t ∈ T \ b K. Then, for those t and kxk

_{∞}

## ≤ M, kyk

∞## ≤ M , we get

## kSx(t) − Sy(t)k

## ≤

\

K

## kG(t, s)k · kr(x(s), s) − r(y(s), s)k ds

## +

\

T \K

## kG(t, s)k(kr(x(s), s) − b(s)k + kr(y(s), s) − b(s)k) ds

## ≤ (ε/(4T )) · 2T + 2L · (ε/(4L)) = ε.

## Theorem 4. Let G and r satisfy the assumptions of the previous theo- rem. Moreover, suppose that

## (2) R = lim sup

kxk→∞

## sup

t∈T

## kr(x, t)k

## kxk < 1/L where L = sup

_{t∈T}

T

T

## kG(t, s)k ds. Then the integral equation Sx = x has a solution in the space BC(T, X).

## P r o o f. By assumption (2), for ε = 1/L − R > 0, choose M > 0 such

## that kr(x, t)k/kxk < R + ε = 1/L for any t ∈ T and kxk > M . Deﬁne

## T = sup

_{kxk≤M, t∈T}

## kr(x, t)k. Using Lemma 4, we get T < ∞. Hence, for

## kxk

∞## ≤ max{LT, M }, kSxk

∞## ≤ sup

t∈T

^{\}

{s:kx(s)k≤M }

## kG(t, s)k · kr(x(s), s)k ds

## +

\

{s:kx(s)k>M }

## kG(t, s)k · kr(x(s), s)k ds

## ≤ sup

t∈T

^{\}

{s:kx(s)k≤M }

## kG(t, s)k max{T, M/L} ds

## +

\

{s:kx(s)k>M }

## kG(t, s)k(max{LT, M }/L) ds

## = sup

t∈T

\

T

## kG(t, s)k max{T, M/L} ds

## ≤ L max{T, M/L} = max{T L, M }.

## Then S : B(0, max{T L, M }) → B(0, max{T L, M }) and, by the Schauder theorem, using Theorem 3, we obtain a ﬁxed point for S.

## Corollary. Let T be a closed cone in a Banach space and let r : X × T → X satisfy conditions (v), (vi) of Theorem 3 and condition (2). Let G : T × T → L(X) be of the form G(t, s) = H(t − s) for any t, s ∈ T , where H : T − T → L(X) (− stands for algebraic difference here) is a given continuous function such that the integral

T

T −T

## kH(t)k dt is finite. Then the integral equation Sx = x has a solution in the space BC(T, X).

## 4. Applications to Wiener–Hopf equations. The last corollary from the previous section can be easily applied to nonlinear Wiener–Hopf equa- tions on a half-line (T = [0, ∞)) or on a closed half-space (T = R

^{k−1}

## × [0, ∞)), which will be illustrated by the following theorems.

## Theorem 5. Consider the equation (3)

∞

\

0

## H(t − s)r(x(s), s) ds = x(t)

## where H : R → R is continuous and integrable, and r : R × [0, ∞) → R is continuous. Moreover , if r satisfies (vi) of Theorem 3 and

## lim sup

|x|→∞

## sup

t∈[0,∞)

## |r(x, t)|

## |x| < 1

## L where L =

∞

\

−∞

## |H(t)| dt, then equation (3) has a bounded and continuous solution x.

## Remark. The conditions imposed on r are, in particular, satisﬁed if

## r(x, t) = f (x)c(t)+b(t) where f : R → R and c, b : [0, ∞) → R are continuous

## and satisfy lim

t→∞## kc(t)k = 0 and lim

kxk→∞## kf (x)k/kxk = 0.

## Theorem 6. Let R

^{k}

_{+}

## = {t = (t

1## , . . . , t

k## ) ∈ R

^{k}

## : t

k## ≥ 0}. Consider the equation

## (4)

\

R^{k}

+

## H(t − s)r(x(s), s) ds = x(t)

## where r : R × R

^{k}

_{+}

## → R and H : R

^{k}

## → R are continuous. Moreover , if r satisfies (vi) of Theorem 3 and

## lim sup

|x|→∞

## sup

t∈[0,∞)

## |r(x, t)|

## |x| < 1

## L where L =

\

R^{k}

## |H(t)| dt < ∞, then equation (4) has a bounded and continuous solution x.

## Remark. The conditions imposed on r are satisﬁed if, in particular, r(x, t) = f (x)c(t) + b(t) where f : R → R and b, c : R

^{k}

_{+}

## → R are continuous and satisfy lim

t→∞## |c(t)| = 0 and lim

|x|→∞## |f (x)|/|x| = 0.

## 5. Applications to ODE’s. Let T = R and let X be a Banach space.

## Assume that r : X × R → X is continuous and A ∈ L(X) is such that Sp(A) ∩ {α ∈ R : α ≤ 0} = ∅. Consider the following boundary value problem in X:

## (5) x

^{′′}

## = Ax + r(x, t),

## x bounded on R.

## The boundedness of solutions plays here the role of a boundary condition.

## Under some additional assumptions on r, problem (5) will be transformed to an integral equation of Hammerstein type and, for the latter, Theorem 4 will be applied to obtain the existence of solutions.

## First, let us recall the notion of the Green function for problem (5).

## The Green function for (5) is a function G : R × R → L(X) satisfying the following conditions:

## (i) For any s ∈ R, G(·, s) satisﬁes the linear equation, i.e.

^{∂}

_{∂t}

^{2}

^{G}2

## (t, s) = AG(t, s) for any t 6= s,

## (ii) lim

_{s→t}−

∂G

∂t

## (t, s) − lim

_{s→t}

^{+}

^{∂G}

_{∂t}

## (t, s) = I for any t ∈ R,

## (iii) G(·, s) satisﬁes the boundary condition for any s ∈ R (i.e. is bounded on R).

## If we have such a function it is easy to see that any solution of the integral equation

## (6) x(t) =

∞\

−∞

## G(t, s)r(x(s), s) ds

## satisﬁes the boundary value problem (5).

## Indeed, integrating and using the properties of the Green function, we obtain

## x

^{′′}

## (t) = ∂

## ∂t

t\−∞

## ∂G

## ∂t (t, s)r(x(s), s) ds + ∂

## ∂t

∞\t

## ∂G

## ∂t (t, s)r(x(s), s) ds

## =

∞

\

−∞

## ∂

^{2}

## G

## ∂t

^{2}

## (t, s)r(x(s), s) ds +

s→t

## lim

^{−}

## ∂G

## ∂t (t, s) − lim

s→t^{+}

## ∂G

## ∂t (t, s)

## r(x(t), t)

## = Ax(t) + r(x(t), t).

## The Green function for (5) will now be constructed. Using the Dunford operational calculus from [6, pp. 225–228], we can deﬁne the operators

## A

^{−1/2}

## = − 1 2πi

\

Γ

## λ

^{−1/2}

## (A − λI)

^{−1}

## dλ,

## −A

^{1/2}

## = 1 2πi

\

Γ

## λ

^{1/2}

## (A − λI)

^{−1}

## dλ,

## exp(−A

^{1/2}

## ) = − 1 2πi

\

Γ

## exp(−λ

^{1/2}

## )(A − λI)

^{−1}

## dλ,

## where Γ is a Jordan curve around Sp(A) with counterclockwise orientation.

## Now, we deﬁne

## G(t, s) = − 1

## 2 A

^{−1/2}

## exp(−A

^{1/2}

## |t − s|).

## From [6, Corollary 3, pp. 245–246] one gets

## ∂

^{2}

## G

## ∂t

^{2}

## (t, s) = − 1

## 2 A

^{1/2}

## exp(−A

^{1/2}

## (t − s)) = AG(t, s) for t 6= s.

## Since Sp(−A

^{1/2}

## ) ⊂ {α ∈ C : Re α < 0}, from [2, Theorem 4.1, pp. 42–43]

## we get

## (7) kG(t, s)k ≤ N e

^{−v|t−s|}

## for some constants N, v > 0. This implies the boundedness of the function G(·, s) for any s ∈ T . Finally, from [6, Theorem, p. 226] and [6, Corollary 3, pp. 245–246] we obtain

s→t

## lim

^{−}

## ∂G

## ∂t (t, s) − lim

s→t^{+}

## ∂G

## ∂t (t, s) = lim

s→t^{−}

## ∂

## ∂t −

^{1}

_{2}

## A

^{−1/2}

## exp(−A

^{1/2}

## (t − s))

## − lim

s→t^{+}

## ∂

## ∂t −

^{1}

_{2}

## A

^{−1/2}

## exp(−A

^{1/2}

## (s − t))

## = lim

s→t^{−}
1

2

## exp(−A

^{1/2}

## (t − s)) − lim

s→t^{+}

## −

^{1}

_{2}

## exp(−A

^{1/2}

## (s − t))

## =

^{1}

_{2}

## I +

^{1}

_{2}

## I = I.

## We have thus shown that G is the Green function for problem (5) and any solution of the integral equation (6) satisﬁes the boundary value problem (5). Now, we are ready to formulate the existence theorem for (5).

## Theorem 7. Assume that A ∈ L(X) is such that Sp(A) ∩ {α ∈ R : α ≤ 0} = ∅ and r(x, t) = c(t)f (x) + b(t) where c : R → R is continuous, lim

|t|→∞## |c(t)| = 0, b : R → X is continuous and bounded, and f : X → X is completely continuous. Then BVP (5) has at least one solution provided

## (8) lim sup

kxk→∞

## kf (x)k

## kxk < v 2N kck

∞## .

## P r o o f. To prove the existence of solutions for the integral equation (6), we have to show that the integral operator S deﬁned by the right-hand side of (6) satisﬁes the assumptions of Theorem 4.

## First, we show that the assumptions of Theorem 3, implying the complete continuity of S, are satisﬁed.

## From inequality (7) it follows that conditions (ii)–(iv) of Theorem 3 are satisﬁed. Assumption (v) is also satisﬁed since the functions b, c are continuous and f maps bounded subsets of X into relatively compact ones.

## The last assumption (vi) is satisﬁed since

t→∞

## lim sup

kxk≤M

## kr(x, t) − b(t)k = lim

t→∞

## sup

kxk≤M

## kc(t)f (x)k

## ≤ sup

kxk≤M

## kf (x)k lim

t→∞

## |c(t)| = 0.

## Now, we show that condition (2) of Theorem 4 is satisﬁed. From inequal- ity (7) we get

## L = sup

t∈R

∞\

−∞

## kG(t, s)k ds ≤ N sup

t∈R

^{t}

^{\}

−∞

## e

^{−v(t−s)}

## ds +

∞\

t

## e

^{−v(s−t)}

## ds

## = 2N v . Then

kxk→∞

## lim sup

t∈T

## kc(t)f (x) + b(t)k

## kxk ≤ kck

∞## lim

kxk→∞

## kf (x)k kxk < v

## 2N = 1 L . From Theorem 4 we get the assertion.

## Analogously we obtain the existence of a solution of the boundary value problem on a half-line.

## Let T = [t

0## , ∞) and let X be a Banach space. Assume that r : X ×

## [t

0## , ∞) → X is continuous and A ∈ L(X) satisﬁes Sp(A) ∩ {α ∈ R :

## α ≤ 0} = ∅. Consider the boundary value problem x

^{′′}

## = Ax + r(x, t), x(t

0## ) = 0,

## (9)

## x bounded on [t

0## , ∞).

## Deﬁne

## G(t, s) =

^{1}

_{2}

## A

^{−1/2}

## ((exp(−A

^{1/2}

## (s + t − 2t

0## )) − exp(−A

^{1/2}

## |t − s|)).

## From [2, Theorem 4.1, pp. 42–43] we get

## (10) kG(t, s)k ≤ N

1## e

^{−v(t+s−2t}

^{0}

^{)}

## + N

2## e

^{−w|t−s|}

## for some constants N

1## , N

2## , v, w > 0. As in the previous case, one can verify that G is the Green function for problem (9), so any solution of the integral equation

## (11) x(t) =

∞\

t0

## G(t, s)r(x(s), s) ds

## satisﬁes the boundary value problem (9). Now, we can state

## Theorem 8. Assume that A ∈ L(X) is such that Sp(A) ∩ {α ∈ R : α ≤ 0} = ∅ and r(x, t) = c(t)f (x)+b(t) where c : [t

0## , ∞) → R is continuous, lim

_{|t|→∞}

## kc(t)k = 0, b : [t

0## , ∞) → X is continuous and bounded, and f : X → X is completely continuous. Then BVP (9) has at least one solution provided

## (12) lim sup

kxk→∞

## kf (x)k

## kxk < vw

## kck

∞## (N

1## w + N

2## v) .

## P r o o f. From inequality (10) we see that assumptions (ii)–(iv) of The- orem 3 are satisﬁed. As in the previous theorem assumptions (v), (vi) are satisﬁed, so the operator S deﬁned by the right-hand side of (11) is com- pletely continuous.

## Now, we show that it satisﬁes condition (2) of Theorem 4. By inequality (10),

## L = sup

t∈[t0,∞)

∞\

t0

## kG(t, s)k ds ≤ sup

t∈[t0,∞)

∞\

t0

## (N

1## e

^{−v(t+s−2t}

^{0}

^{)}

## + N

2## e

^{−w|t−s|}

## ) ds

## = sup

t∈[t0,∞)

## − N

1## v e

^{−v(t+s−2t}

^{0}

^{)}

## |

^{∞}

_{s=t}

_{0}

## + N

2## w e

^{−w(t−s)}

## |

^{t}

_{s=t}

_{0}

## − N

2## w e

^{−w(s−t)}

## |

^{∞}

_{s=t}

## = sup

t∈[t0,∞)

## N

1## v e

^{−v(t−t}

^{0}

^{)}

## + N

2## w − N

2## w e

^{−w(t−t}

^{0}

^{)}

## + N

2## w

## = N

1## v + N

2## w ,

## whence, by (12), lim sup

kxk→∞

## sup

t∈T

## kc(t)f (x) + b(t)k

## kxk ≤ kck

∞## lim sup

kxk→∞